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Current time:0:00Total duration:13:25

2003 AIME II problem 10

Video transcript

two positive integers differ by 60 the sum of their square roots is the square root of an integer that is not a perfect square what is the maximum possible sum of the two integers so the second sentence is a little confusing let's take it step by step so two positive integers differ by 60 so let's say we have their a and B so we have a and B they're both greater than zero they're also both integers so they're also they're also both integers and they differ by 60 so a minus B is equal to 60 or we could add B to both sides and we get a is equal to B plus a 60 there's 60 apart fair enough now this second sentence the sum of their square roots so the sum of the square roots square root of a plus square root of B the sum of the square roots is the square root of an integer that is not a perfect square so it's equal to the square root of some integers so C is an integer C is an integer but it's not a perfect square so square root of C not is not an integer so that's the second part what is the maximum possible sum of the two integers so if we can maximize B and and find an a that fits it or an A it'd be that meet this constraint smoking them as big as possible then the sum of a and B will also be maximized so let's think about that in a second but just to simplify this is actually the really confusing part because C is an integer and square root of C is not an integer and to make it a little less confusing let's just square both sides of this equation over here so square root of a squared is so we're just going to square it so this we're just gonna square both sides let me just we're just going to square both sides of this equation so this is going to be square root of a times square root of a which is a plus 2 times the square root of a times the square root of B which is 2 times the square root of a B plus the square root of B squared which is just going to be B is equal to is equal to C now we know that a is an integer we know that V is an integer and we know that C is an integer we know that these are integers so if I was an integer plus something plus an integer equaling another integer this thing in the middle has to be an integer this thing in the middle has to be has to be an integer actually I'll show you right now that not only does this thing have to be an integer square root of a B has to be integer and you might be saying wait wait wait wait a second can't I just set if this just thing just has to be an integer can't I just say let me do it over here on the side can't I just say that two times the square root of a B has to be equal to some integer and you could divide both sides by two and you could say well maybe the square root of a B is equal to K over 2 where this is not an integer where this is not where this is not integer I claim that this thing does have to be an integer but you say hey look just straight from the math this could be an integer but maybe you divide both sides by 2 and this doesn't have to be an integer and if this is not an integer if this is not an integer that means that K is odd because otherwise it would be divisible by 2 now if we square both sides of this we would get a B is equal to K squared over 4 well this is going to be an odd times an odd it's going to be an odd squared so this is still going to be odd which is not divisible by 4 so this thing on the right hand side this thing on the right hand side will still not be an integer it will not be an integer but remember a and B are integers so a times B this is an integer this is an integer so if a and B are integers there's no way that the square root of its product can be something divided by 2 where this doesn't reduce anymore where this doesn't simplify or some odd number divided by 2 so we just get a contradiction here or another way to say it is the square root of a B because because this whole thing has to be an integer and both a and B are integers the square root of a B has to be has to be an integer has to be an integer but if the square root of a B has to be an integer then that means that a B has to be a perfect square so that I'll start over here so I don't run a space that means that a B actually let me do it right under that because it's a straight result so that means that a times B is a perfect square is a perfect that means that a times B is a perfect square so let's see how we can deal with this a little bit we can actually do a substitution for a a is equal to B plus 60 so this means that B plus 60 times B is equal to a perfect square so it could be equal to some number n N squared where n is an integer where n is an integer and let's see if we can play around with this a little bit put some constraints on what we're doing so if you multiply this out you get B squared plus 60 B B squared plus 60 B I'll leave some space over here is equal to N squared let's see we can complete the square over here and maybe we can subtract this over and we'll have a difference of squares so we want to complete the square here you take half of this so half of 60 the half of the coefficient that's 30 and square it so this would be plus 900 - 900 we haven't changed it we're adding and subtracting the same thing but what that does for us is that this part over here becomes B plus 30 B plus 30 squared and then we have the minus 900 is equal to N squared and then we could subtract N squared from both sides we could subtract N squared from both sides and add 900 to both sides and we get B plus 30 and you really just have to play around with this thing and see what constraints you can get v plus 30 squared minus N squared is equal to 900 and this is just a difference of squares so this can be factored this is V plus 30 this is a plus B or I shouldn't use a and B but it's the square root of this plus the square root of this so plus n times the square root of this minus the square root of this so B plus 30 B plus 30 minus n so we can factor it just like that and that's about as good of a constraint as we can get remember we want to be able to maximize B if we're maximizing VA is only 60 above B so we're also going to be maximizing a and so we'll also be maximizing the sum so we want B to be as high as possible as high as possible so B plus 30 plus n is going to be some factor of 900 and B plus 30 minus n will be the other factor of 900 and we can actually do a little bit more of an insight here we have two factors their product is an even number so it's an even product even even product and if we were to take the sum of these two factors if you were to take V plus 30 plus N and add it to B plus 30 minus n so we're just taking the sum of these two factors we get to be plus 60 and these cancel out and so this is also even so if I take two numbers and their product is even and their sum is also even that means that two numbers have to be even so both this guy and this guy over here have to be even both of them have to be even and at this point I think we've kind of used up all the information in the problem we really just have to try out factors of 900 that meet these constraints where both are even and let's try with the most extreme things and attempt to maximize in an attempt to maximize our B so let's think about the factors of 900 you could do 1 and 900 but remember both factors have to be even one is not even weakened truck we could try 2 and 450 where this will be the higher factor this will be the lower factor because we're subtracting a number here so let's try that out maybe B actually let me do it over here maybe I'll do it actually I'll do it over here I'll do maybe this is 450 and this is 2 so we have B plus 30 plus n is equal to 450 and that B plus 30 minus n is equal to 2 and now we can solve for B you can add these two equations you get 2b plus 60 these cancel out is equal to 450 to subtract 60 from both sides to be is equal to 392 or B is equal to 196 so let's see how this works out if B is equal to 196 so if B is equal to 196 then a is just 60 Plus that so a is equal to 200 a is equal to 256 and let's see what it does to this equation over here square root of a is 16 so we'd have the square root of 256 plus the square root of 196 well that's equal to 16 plus 14 which is equal to 30 so it looks like it might work but remember this thing over here the square root of C is not a perfect square this value right over here that's the square root of C that is not or I should say the square root of C is not an integer it cannot be an integer if it was then C which in this case would be 900 would be a perfect square but it they told us of a problem C is not a perfect square so this one actually does not work because this would give us a C that is a perfect square it gives us 900 square root of C is 30 so we have to cross these out too although there's very tantalizing solution since it got gave us nice clean numbers so let's keep trying factors of 900 so remember you could try 3 and 300 but we know that they both have to be even so we could just reject this because 3 is not even let's so maybe it's 4 and 4 goes into 900 125 times you might be tempted to try these out but 125 is not even so we could keep going so then we could go to 5 well once again 5 is not even so that won't work we could try 6 6 goes into 900 it goes into 90 15 times so it goes into 900 150 times they're both even numbers so maybe just maybe these will work so let's try it out again so let me scroll down a little bit so you could have B plus 30 plus and that would be the larger number is equal to 150 because we're adding the end there and then you have B plus 30 minus n would be equal to 6 you add both equations you get 2 B plus 60 is equal to 100 and is equal to 156 or subtracting 60 from both sides to be is equal to let's see if you subtract this is equal to 96 or V is equal to 48 so here's our next candidate our next candidate is V is equal to 48 then a would be 60 more than this a would be equal to 108 60 more than that and if I were to take the square root of 108 plus the square root of 48 what does that give me what does that give me 108 is 36 times 3 so this is the square root of 36 times the square root of 3 or 6 square roots of 3 plus this is 16 times 3 so it's going to be the square root of what 48 is 16 times 3 so it's the square root of 16 times the square root of 3 so it's 4 square roots of 3 so when you add these two together you get 10 square roots of 3 so this fits our bill for the square root of C so over here we have the square root of C is equal to 10 roots of 3 it's definitely not an integer so it makes that and if you square it if you square it if you square it you would get C as being equal to 100 times 3 which is equal to 300 and it's an integer so it works so our best shot our best values for a and B V is 48 a is 108 if we take the sum 108 plus 48 we get 16 5 and 156 and you could see when you keep trying obviously we've had the best result when we had when we had B being 196 but as we go as we get our factors closer and closer together the B is going to go lower and lower lower and we're obviously going to have a less of a maximum value for a plus B so this is as we got from the most extreme factors of 900 we keep going until something met this met our constraint if we kept going if we tried I don't know if we tried 8 and I don't know if that would even work as a factors but if we just tried it kept looking at the other factors we would have gotten we actually we've got numbers that are closer to each other and that would have led to lower values for our B so this right here is our maximum be the 48 max or the the the B that will maximize our result and the sum is 156